A force vecF = (3xy -5z)hatj + 4zhatk is applied on a particle. The work done by the force when the particle moves from point (0,0,0) to point (2,4,0) along the path y = x^2 is?

1 Answer
MetaPhysik
Feb 17, 2018

W=48 [J]

Explanation:

Use the definition of work,

W = intvecf*dvecr

Instead using unit vector notations, let's use coordinate to represent the vectors, i.e., vec f = (0, 3xy -5z, 4z); and dvecr = (dx, dy, dz)

W = int_("(0,0,0)" )^("(2,4,0)")(0, 3xy -5z, 4z)*(dx, dy, dz)#

W = int_("(0,0,0)" )^("(2,4,0)") (3xy -5z)dy +4zdz
W= int_0^4 (3xy -5z)dy + int_0^0 4zdz
because no displacement in z, z-component force contribute no work, hence

W= [3/2xy^2-5zy]_0^4 + 0

W= 24x-20z
The above is the work done alone the path y=x^2 on a z-plane.

At point (2,4, 0),

W= 24x-20z= 24(2)-20(0)=48

The units are not specified, hence the work done has not unit, otherwise it is assumed as W=48J

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